Question: You have found the following ages (in years) of all 4 tigers at your local zoo: $ 13,\enspace 8,\enspace 11,\enspace 4$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{13 + 8 + 11 + 4}{{4}} = {9\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $13$ years $4$ years $16$ years $^2$ $8$ years $-1$ years $1$ year $^2$ $11$ years $2$ years $4$ years $^2$ $4$ years $-5$ years $25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{16} + {1} + {4} + {25}} {{4}} $ $ {\sigma^2} = \dfrac{{46}}{{4}} = {11.5\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{11.5\text{ years}^2}} = {3.4\text{ years}} $ The average tiger at the zoo is 9 years old. There is a standard deviation of 3.4 years.